KindiakMath

Proving Volumes

In this post, we use integral calculus to prove, or at least derive, several famous formulae in geometry.

Definition 1. The volume of a prism with constant base area A and height h is defined by Ah.

Corollary 1. The volume of a cylinder with base radius r and height h is \pi r^2 h.

We can use the prism to derive the volume of a cone (more generally, a pyramid).

Definition 2. Let A(x) be a continuous function defined on [a , b] that denotes the cross-sectional area of a solid at x. Define x_k = a + (b-a)k/n and \Delta x = (b-a)/n. Its volume can be computed using the limit

\displaystyle V := \lim_{n \to \infty} \sum_{k=1}^n A(x_k) \Delta x = \int_a^b A(x)\,\mathrm dx.

Theorem 1. The volume of a pyramid with base area A and height h is \frac 13 Ah.

Proof. Using similar triangles and similar areas, we can consider the continuous function A(x) = Ax^2/h^2 on [0, h]. Using the integral definition of volume,

V = \displaystyle \int_0^h A(x)\, \mathrm dx = \int_0^h \frac{Ax^2}{h^2}\, \mathrm dx = \frac{A}{h^2} \int_0^h x^2\, \mathrm dx = \frac{A}{h^2} \cdot \frac{h^3}{3} = \frac 13 Ah.

Corollary 2. The volume of a cone with base radius r and height h is \frac 13 \pi r^2h.

Proof. Contextualise the previous result to A = \pi r^2.

In fact, whenever A(x) = \pi (f(x))^2 corresponds to a circle with radius f(x), we obtain the volume of revolution formula:

Theorem 2 (Volume of Revolution). Let f(x) \geq 0 be a continuous function on [a, b]. The volume of revolution V of f(x) about the x-axis is

\displaystyle V = \pi \int_a^b (f(x))^2\, \mathrm dx.

We can also prove the formula for the volume of a sphere using a similar integration technique.

Theorem 3. The volume of a sphere with radius r is \frac 43 \pi r^3.

Proof. Recall that the upper-half semicircle with radius r has equation y = \sqrt{r^2 -x^2}. Consider the function A(x) = \pi y^2 = \pi (r^2-x^2) defined on [-r, r]. Using the integral definition of volume,

\displaystyle \begin{aligned} \int_{-r}^r A(x)\,\mathrm dx &= \pi \int_{-r}^r (r^2-x^2)\, \mathrm dx \\ &= \pi \left[ r^2x - \frac{x^3}{3} \right]_{-r}^r \\ &= \pi \left(2r^3 - \frac{2r^3}{3}\right) = \frac{4\pi r^3}{3}. \end{aligned}

What about the surface area of a sphere? We first need the surface area of a cone.

Lemma 1. The surface area of a cone with slant height l and radius r is given by \pi r l.

Proof. The cone can be thought of as a sector with radius l and arc length 2 \pi r, yielding an area of

\displaystyle \frac{ 2 \pi r }{ 2 \pi l } \cdot \pi l^2 = \pi r l.

Remark 1. The angle that this sector corresponds to is 2\pi r/l.

Next, we require the surface area of a frustum.

Lemma 2. The surface area of a frustum with radii r < R and slant height l is given by \pi (r + R) l.

Proof. Let the total slant height of the completed cone be s + l. By similar triangles,

\displaystyle \frac{r}{s} = \frac{R}{s+l} \quad \iff \quad s = \frac{rl}{R-r}.

By subtracting surface areas of the smaller cone from the larger cone, we have a surface area of

\displaystyle \pi R(s+l) - \pi rs = \pi (R-r)s + \pi Rl = \pi rl + \pi Rl = \pi(r+R)l.

This helps us define the surface area of revolution.

Definition 3. Let f(x) \geq 0 be a continuous function on [a, b]. The surface area of revolution S is defined by

\displaystyle S := \lim_{n \to \infty} \sum_{k=1}^n \pi (r_k+R_k) l_k,

where r_k = f(x_{k-1}), R_k = f(x_k), l_k = \sqrt{(R_k - r_k)^2 + \Delta x^2}.

Since r_k+R_k \to 2f(x_k) and l_k \to \sqrt{1+(f'(x_k))^2}\Delta x, taking n \to \infty yields

\displaystyle S = \int_a^b 2 \pi f(x) \sqrt{1+(f'(x))^2}\, \mathrm dx.

Setting y = \sqrt{r^2-x^2} and computing integrals therefore yields the surface area of a sphere.

Theorem 4. The surface area of a sphere with radius r is given by 4 \pi r^2.

—Joel Kindiak, 31 Oct 24, 0229H

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